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MCAT Question A Day - 3/11/14 - Answer!

What is the average power output of a 70-kg woman who runs up an incline to a height of 2 meters in 2 seconds?

A.  343 W
B.  686 W
C.  1372 W
D.  1715 W

The correct answer is (B). Pavw/t, but the work done (w) is mgh; therefore,

 Pav = (70 kg)(9.8 m/s2)(2m) / (2s)  =  686 kg m2/s3 = 686 J/s = 686 W

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